题解

#include <iostream>
#include <iomanip>
using namespace std;

int main() {
    int n, k;
    cin >> n >> k;

    // 计算A类（k的倍数）
    int m = n / k; // A类数的个数
    long long sumA = (long long)k * m * (m + 1) / 2;
    double avgA = (double)sumA / m;

    // 计算B类（非k的倍数）
    long long total = (long long)n * (n + 1) / 2;
    long long sumB = total - sumA;
    int countB = n - m;
    double avgB = (double)sumB / countB;

    // 输出，保留1位小数
    cout << fixed << setprecision(1) << avgA << " " << avgB << endl;

    return 0;
}